11.2.3 Dedekind reals are Dedekind complete

We obtained ${ℝ}_{𝖽}$ as the type of Dedekind cuts on $\mathbb{Q}$. But we could have instead startedwith any archimedean ordered field $F$ and constructed Dedekind cuts on $F$. These wouldagain form an archimedean ordered field $\overline{F}$, the Dedekind completion of $F$,with $F$ contained as a subfield. What happens if we apply this construction to${ℝ}_{𝖽}$, do we get even more real numbers? The answer is negative. In fact, we shall prove astronger result: ${ℝ}_{𝖽}$ is final.

Say that an ordered field $F$ is admissible for $\mathrm{\Omega }$when the strict order on $F$ is a map .

Theorem 11.2.1.

Every archimedean ordered field which is admissible for $\mathrm{\Omega }$ is a subfield of ${\mathrm{R}}_{\mathrm{d}}$.

Proof.

Let $F$ be an archimedean ordered field. For every $x:F$ define $L,U:\mathbb{Q}\to \mathrm{\Omega }$ by

(We have just used the assumption that $F$ is admissible for $\mathrm{\Omega }$.)Then $(L_x,U_x)$ is a Dedekind cut. Indeed, the cuts are inhabited and rounded because$F$ is archimedean and is transitive, disjoint because is irreflexive, andlocated because is a weak linear order. Let $e:F\to {ℝ}_{𝖽}$ be the map $e(x):\equiv (L_x,U_x)$.

We claim that $e$ is a field embedding which preserves and reflects the order. First ofall, notice that $e(q)=q$ for a rational number $q$. Next we have the equivalences,for all $x,y:F$,

so $e$ indeed preserves and reflects the order. That $e(x+y)=e(x)+e(y)$ holdsbecause, for all $q:\mathbb{Q}$,

The implication from right to left is obvious. For the other direction, if then there merely exists $r:\mathbb{Q}$ such that , and by taking $s:\equiv q-r$ we get the desired $r$ and $s$. We leave preservation of multiplication by $e$ asan exercise.∎

To establish that the Dedekind cuts on ${ℝ}_{𝖽}$ do not give us anything new, we need just onemore lemma.

Lemma 11.2.2.

If $F$ is admissible for $\mathrm{\Omega }$ then so is its Dedekind completion.

Proof.

Let $\overline{F}$ be the Dedekind completion of $F$. The strict order on $\overline{F}$ isdefined by

Since $U(q)$ and $L^{\prime }(q)$ are elements of $\mathrm{\Omega }$, the lemma holds as long as $\mathrm{\Omega }$is closed under conjunctions and countable existentials, which we assumed from the outset.∎

Corollary 11.2.3.

The Dedekind reals are Dedekind complete: for every real-valued Dedekind cut $\mathrm{(}L\mathrm{,}U\mathrm{)}$there is a unique $x\mathrm{:}{\mathrm{R}}_{\mathrm{d}}$ such that and for all $y\mathrm{:}{\mathrm{R}}_{\mathrm{d}}$.

Proof.

By lem:cuts-preserve-admissibility the Dedekind completion ${\overline{ℝ}}_{𝖽}$ of ${ℝ}_{𝖽}$is admissible for $\mathrm{\Omega }$, so by RD-final-field we have an embedding ${\overline{ℝ}}_{𝖽}\to {ℝ}_{𝖽}$, as well as an embedding ${ℝ}_{𝖽}\to {\overline{ℝ}}_{𝖽}$. But these embeddings must beisomorphisms, because their compositions are order-preserving field homomorphisms whichfix the dense subfield $\mathbb{Q}$, which means that they are the identity. The corollary nowfollows immediately from the fact that ${\overline{ℝ}}_{𝖽}\to {ℝ}_{𝖽}$ is an isomorphism.∎